okay great, all of the previous answers are good but yours made the most sense to me. If i am interpreting you correctly, then to "fix" e to the end, are you essentially saying you are making an 8 letter word "committe" instead?
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ylun.caFeb 18 '14 at 2:01

Yes, you are making a eight letter word committe which is different from a 9 letter word not to end in "e"
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satish ramanathanFeb 18 '14 at 2:05

Both you and Andre's answers are great but yours is a more simple process to follow. Thank you.
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ylun.caFeb 18 '14 at 2:21

There are $9$ slots to be filled. An e cannot be at the end, so we can pick the pair of locations occupied by the e's in $\binom{8}{2}$ ways. For each of these, the locations of the m's can be picked in $\binom{7}{2}$ ways. Now the locations of the t's can be chosen in $\binom{5}{2}$ ways. The remaining $3$ slots can be filled in $3!$ ways. This gives a total of $\binom{8}{2}\binom{7}{2}\binom{5}{2}3!$ words.

Remark: It is very useful to place the e's first. The order in which we consider the rest does not matter. For example, after the e's are placed, the c can be placed in $\binom{7}{1}$ ways. For each of these, the t's can be placed in $\binom{6}{2}$ ways. and so on.

For any word how many permutations give the same word? The letters $t$ can be switched (2 ways), the letters $m$ can be switched (2 ways) and the letters $e$ can also be switched (2 ways) therefore for any word there are $2\cdot2\cdot2=8$ permutations giving the same word.

This is problem 3 of chapter 1 from Lászlo Lovász book Combinatorial Problems and Excercises. He uses the word caharacterization. Here I copy what he says about the word characterization.

There are 16! permutations of the word CHARACTERIZATION however not all of these give new words; in fact,in any permutation, if we exchange the three $A$'s,the two $C$'s,the two $R$'s,the two $I$'s or the two $T$'s we get the same word. Thus for any permutation, there are $3!\cdot2\cdot2\cdot2\cdot=96$ permutations which give the same word, so the result is $\frac{16!}{96}$

In General, if there are $k_A$ $A$'s, $k_b$ $B$'s etc. then the result is